Let $k$ be an algebraically closed field. Consider the curve $V(y^2 = x^3 - 3x^2 + 2x) = C \subset \mathbb{A}^2$.
Now, we know that its the zero locus of some ideal, from what I gather this ideal is principal but im not able to see this (how can I see this?).
Given that $C = V(I)$ for a principal ideal $I=(f)$, I want to consider its projective closure in $\mathbb{P}^2$ for $U_i = \{ (a_0:a_1:a_2) \in \mathbb{P}^2 : a_i \not= 0 \}$. Let's take $i=2$ for example, then $\overline{\psi_2(C)} \subset \mathbb{P}^2$ is closed subset defined by the homogeneous ideal generated by $\text{hom}_2(g)$ for $g \in I$. Note that we define $\psi_i: \mathbb{A}^n \to U_i$ by $(a_1,...,a_n) \mapsto (a_1:...:a_{i-1}:1:a_{i+1}:...:a_n)$ (when working over $\mathbb{A}^n$) which is a homeomorphism. So we are considering the closure of the curve $C$ under this homeomorphism.
Here I have a second question. Now, something tells me that this projective closure is just the zero locus of the homogenized polynomial $X^3 - 3X^2Z + 2XZ^2 - ZY^2$, but I am stuck trying to show this.